91Ó°ÊÓ

Given,

An Ideal gas with moles n = 0.305 mol undergoes isothermal compression at temperature $T=22C°$or $295.15K°$. And the work done is negative, Work done W = -392J .The final pressure$p_2=1.76atm$ . Work done when pressure changes is given by:

$W{∫}_{V_1}^{V_2}pdV$

From Ideal gas law we know that, $p=\frac{nRT}{v}$

Now, substituting the value of p in integration while keeping nRT constant we get:

$$\begin{gathered} W=nRT∫\frac{1}{V}dV \\ =nRT\ln \frac{V_2}{V_1} \end{gathered}$$

Again, from ideal gas law we get: $\frac{V_2}{V_1}=\frac{p_1}{p_2}$

Now, substituting the values of$\frac{V_2}{V_1}$ we get:

$$\begin{gathered} W=nRT\ln \frac{P_1}{P_2} \\ {e}^{\frac{-W}{nRT}}=\frac{P_1}{P_2} \end{gathered}$$

Putting the values in the above equation and solving for$p_1$ :

$$\begin{gathered} p_1=1.76atm.e^{-395J/\left(0.305mol/8.31L/mol.K295.15K\right)} \\ =1.76atm{e}^{-523} \\ =1.76atm\left(0.529\right) \\ =1.042atm \end{gathered}$$

Therefore, the initial pressure is$1.042atm$

As the work done is on the gas, therefore the volume would decrease, hence the pV diagram would be as above.